从给定数组生成增量值数组

Question

如何从给定数组生成增量值数组

这个想法是创建一种菱形形状，一旦到达阵列的中间，阵列的尺寸就开始减小。换句话说，最长的数组将是包含数组中间值的数组或 (array.length/2 + 1)

如果元素不足以完成后半部分的数组，只需将其替换为“E”以指示空白，就像第二个示例一样。

example 1
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p']
//the longest array in length is containing 'i' which is the value at 
    array.length/2 + 1
var output = [
    ['a'],
    ['b','c'],
    ['d','e','f'],
    ['g','h','i','j'], 
    ['k','l','m'],
    ['n','o'],
    ['p']
]

example 2
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t']
enter code here
//the longest array in length is containing 'k' which is the value at array.length/2 + 1
var output = [
    ['a'],
    ['b','c'],
    ['d','e','f'],
    ['g','h','i','j'],
    ['k','l','m','n','o'],
    ['p','q','r','s'],
    ['t','E','E'],
    ['E','E'],
    ['E]
]

这是我尝试过的代码：

const values = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16];
const halfLen = values.length/2 + 1;
var topArr = [];

for(let i = 0; i < values.length; i ++) {
    if(i <= halfLen) {
    topArr.push(values[i])
  }
    
}
console.log(topArr)

var filTopArr = [];

for(let i = 0; i <= topArr.length; i ++) {

    let prevIndex = i - 1;
    
    if(i === 0) {
        filTopArr.push(topArr[i])
    }  else if(i === 1) {
        filTopArr.push(topArr.slice(i, i + i + 1))
    } else {
    filTopArr.push(topArr.slice(i, i + i ))
    }
    
}

console.log(filTopArr)

我的想法是将数组分成两个不同的数组，其中顶部部分的大小将增加，第二/底部部分的大小将减小。

上面的代码有这个输出

[1, [2, 3], [3, 4], [4, 5, 6], [5, 6, 7, 8], [6, 7, 8, 9], [7, 8, 9], [8, 9], [9], []]

Answer 1

一些观察结果：

• 输出中的字符串数量（包括填充“E”字符串）始终是完全平方数（1、4、9、16、25...等）

• 为了知道需要添加多少个“E”字符串，我们需要知道哪个是不小于输入大小的最小完美平方。

• 输出中最长（中间）子数组的大小是该完美平方的平方根。

• 子数组的数量是该数字的两倍减 1。

这导致以下实现：

function diamond(array) {
    // Get least perfect square that is not less than the array length 
    const sqrt = Math.ceil(Math.sqrt(array.length));
    const size = sqrt ** 2;
    // Pad the array with "E" strings so to reach that perfect square size
    const all = [...array, ..."E".repeat(size - array.length)];
    const length = 2 * sqrt;
    return Array.from({length}, (_, width) => {
        return all.splice(0, Math.min(width, length - width));
    }).slice(1); // Skip the first subarray that was produced (empty array)
}

// Demo using the two provided examples:
var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p'];
console.log(diamond(array));

var array = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t'];
console.log(diamond(array));